0 I have a question regarding the quantification of the error in peak integrals due to noise. I have a series of spectra with only one signal. I'm doing a diffusion measurement, so the intensity of this signal decreases, going through the series. I have to quantify the intensity drop, and also the error in this due to low S/N. 1) What would the error in the intensity due to noise be if I would determine the signal intensity from the first point of the FID (as I have only one signal, this is a good method) 2) What would the error be if I first FT the FID, and then determine the integral of the resulting signal (this is easier to do with standard NMR software) I hope someone can help me out! asked Jan 17 '13 at 03:18 Daan 11

 0 The FID contains part of the noise from the entire spectrum in every point, and the first point may or may not indicate the amplitude, depending on the phase. The FT transformed spectrum separates out the noise into the spectral baseline, and corrects the phasing, so the only noise you integrate is that which is at the same frequency as your signal. IMO, you are better off doing the FT before determining the intensity. answered Jan 23 '13 at 11:08 Thanks Jerry. Both in FID as FT each point contains equal noise. In principle, if you phase the FID so that the first point is the highest point, the FT should be nicely phased as well. Of course with a lot of noise, you cannot phase the FID perfectly, but that's also the case with the FT. - Daan (Jan 28 '13 at 04:34) So,the first point of the FID has less noise then the integral of the signal in the FT, and both contain a phasing uncertainty. I'm still not completely sure what would be the better option. This seems to be a classical problem, something must have been written on this I guess. - Daan (Jan 28 '13 at 04:36) The FID contains a portion of all frequencies within the receiver bandwidth in every point. The FT spectrum contains only one very narrow frequency range in each point. - Jerry Hirschinger (Feb 18 '13 at 13:23)